Java RuntimeException:org.hibernate.exception.sqlgrammareException:无法执行查询

我正在使用另一个表联接从一个名为PurchaseDetailContributor的表中检索记录。但我得到的查询错误

下面是我的DAOHibernate.java类，我在其中编写了一个方法

@SuppressWarnings("unchecked")
public List<PurchaseDetailContributor> getPurchaseDetailContributorByCustomerNo(Integer customer_no) {
    System.out.println("The Customer number in PurchaseDetailContributorDAOHIbernate >>>>" + customer_no);
    String sql = " SELECT * FROM purchase_detail_contributor pdc "
             + " INNER JOIN purchasedetail pd ON pdc.purchase_detail_id = pd.purchasedetailid "
             + " INNER JOIN purchase p ON pd.purchasefk = p.purchaseid " 
             + " INNER JOIN user u ON u.userid = p.userid AND u.customer_no = " +customer_no;

    List<PurchaseDetailContributor> pdcList = new ArrayList<PurchaseDetailContributor>();
    Transaction tx = null;
    Session session = getHibernateTemplate().getSessionFactory().openSession();
    try{
        tx = session.beginTransaction();
        SQLQuery sqlQuery = session.createSQLQuery(sql).addEntity(PurchaseDetailContributor.class);
        //sqlQuery.setInteger("customer_no", customer_no);
        pdcList = sqlQuery.list();
        tx.commit();
    }catch (RuntimeException e) {
        System.out.println("getPurchaseDetailContributorByCustomerNo() >>>>- RuntimeException: " + e);
        if (tx != null && tx.isActive()) {
            try {
                // Second try catch as the rollback could fail as well
                tx.rollback();
            } catch (HibernateException e1) {
                System.out.println("Error rolling back transaction: " + e1);
            }
        }
    } finally {
        session.disconnect();
    }

---

sql中没有空格

String sql = "select * from purchase_detail_contributor"
         + "inner join purchasedetail on (purchase_detail_contributor.purchase_detail_id = purchasedetail.purchasedetailid)"
         + "inner join purchase on (purchasedetail.purchasefk = purchase.purchaseid)" 
         + "inner join user on (user.userid = purchase.userid)"
         + "where user.customer_no = :customer_no";

所以sql是

select * from purchase_detail_contributorinner join purchasedetail...

---

这肯定是语法错误

sql中没有空格

String sql = "select * from purchase_detail_contributor"
         + "inner join purchasedetail on (purchase_detail_contributor.purchase_detail_id = purchasedetail.purchasedetailid)"
         + "inner join purchase on (purchasedetail.purchasefk = purchase.purchaseid)" 
         + "inner join user on (user.userid = purchase.userid)"
         + "where user.customer_no = :customer_no";

所以sql是

select * from purchase_detail_contributorinner join purchasedetail...

---

这肯定是语法错误

您发布了正确的代码吗

正如它所显示的错误

11:13:01,074 ERROR [JDBCExceptionReporter:234] Column 'is_approved' not found.

---

你发了正确的代码吗

正如它所显示的错误

11:13:01,074 ERROR [JDBCExceptionReporter:234] Column 'is_approved' not found.

---

添加“内部联接”和“where”子句时需要一些空格，否则sql将类似于“select*from PUSHARE\u detail\u contributorinner join…”，谢谢，现在我按照大家的建议更改了sql查询，但仍然出现以下错误org.hibernate.exception.sqlgrammareexception:无法执行查询您可以在提交sql并将其添加到问题之前打印您的sql吗？我不明白您的意思……请您修改我代码中的错误。以帮助我们理解可能出现什么问题，您可以添加一些信息吗？例如，完整的stacktrace和/或sql变量的值？在添加“内部联接”和“where”子句时，您需要一些空格，否则您的sql将类似于“select*from purchase\u detail\u contributorinner join…”，谢谢，现在我按照大家的建议更改了sql查询，但仍然出现以下错误org.hibernate.exception.sqlgrammareexception:无法执行查询您可以在提交sql并将其添加到问题之前打印您的sql吗？我不明白您的意思……请您修改我代码中的错误。以帮助我们理解可能出现什么问题，您可以添加一些信息吗？例如，完整的stacktrace和/或sql变量的值？创建查询时，必须始终设置参数，如

user.customer\u no=：cno

，然后调用

setParameter（“cno”，customer\u no）

，以防sql注入攻击。这个更安全，因为它是一个整数对象，但是如果你在一个地方这样做，可以安全地假设你在更多地方这样做。创建查询时，你必须始终设置参数，如

user.customer\u no=：cno

，然后调用

setParameter（“cno”，customer\u no）

以防止SQL注入攻击。这一个更安全，因为它是一个整数对象，但是如果你在一个地方这样做，可以安全地假设你在更多地方这样做。